Re-Inventing The Wheel-Part1-Clemente_Figuera-THE INFINITE ENERGY MACHINE

[RMatt]

A few questions about coils/electromagnets.

When several coils are connected in series, does the voltage add together?, does the amperage add together?, does it multipy?

Bob

[RMatt]

Forget last question. Found it on Wikipedia.

"The current through inductors in series stays the same, but the voltage across each inductor can be different. The sum of the potential differences (voltage) is equal to the total voltage. To find their total inductance: L eq=L1+L2+...+Ln"

Bob

[i_ron]

Forget last question. Found it on Wikipedia.

"The current through inductors in series stays the same, but the voltage across each inductor can be different. The sum of the potential differences (voltage) is equal to the total voltage. To find their total inductance: L eq=L1+L2+...+Ln"

Bob

Hi Bob,


Yes that is almost the case, voltage adds and amps stays the same... except it will be less than one coil as the more coils in series the more resistance, so voltage and amps will need to be adjusted accordingly.


Ron

[RMatt]

Thank you for the info Ron.

I recieved 4 of the 3 phase line reactors, and should recieve the other 6 today. ( see Reply# 232 and Reply#234 ). I am going to try using just one for now and reverse one outside coil to get a N-S allignment. I am not going to rewire them until after I have tried them to see if they will work. If they work, I am considering placing all 10 sets ( instead of just 7 sets ) to get a greater output. My main problem will be that I do not have an O-scope, so I will have to find someone with one to run test for me. Hopefully I will have some results by next week.

Bob

[hanon]

Maybe the reason for being a Lenzless generator is simply because both changes in the inducer currents are in opposition. The back emf is a consecuence of the change in the magnetic field, and is calculated by the equation back_emf = -N·S·dB/dt .

Because the current in the N electromagnets is increasing while the current in the S electromagnets is decreasing we can say that their time derivatives are opposite :  dI_N/dt = - dI_S/dt

As the inducer field is calculated as B = (nu·N·I)/length , therefore deriving:  dB/dt = (nu·N/length) · dI/dt . For this reason , with opposite changes in both inducer currents, the change in both induced magnetic fields are  in opposition: dB_N/dt = - dB_S/dt

If both values can be added (I am not sure (please comment..)) dB_final/dt = dB_N/dt + dB_S/dt = 0  . Thus, althought there is a change in the primary magnetic fields (the cause of the induced current), the resulting induced magnetic field (the consecuence) -moving back to the electromagnets- is almost eliminated.

Regards